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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the area of the region $ {(x,y):x^2+y^2 \leq 1 \leq x + y}.$

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Toolbox:
  • If we are given two curves represented by y=f(x);y=g(x),where $f(x)\geq g(x)$ in [a,b],the points of intersection of two curves are given by x=a and x=b,by taking common values of y from the equation of the two curves.
Step 1:
Given : $R=\{(x,y) : x^2+y^2\leq 1\leq x+y\}$
$\Rightarrow R=R_1\cap R_2$
$\Rightarrow R_1=\{(x,y):(x^2+y^2\leq 1\}$
$\Rightarrow R_2=\{(x,y): 1\leq x+y\}$
$x^2+y^2\leq 1$ represents the equation of the circle with centre (0,0) and radius 1.
$x+y=1$ is the equation of a straight line .
Now let us find the points of intersection of these two curve.
$x+y=1$
$\Rightarrow y=1-x$
$y^2=(1-x)^2$
Hence $x^2+(1-x)^2=1$
$\Rightarrow x^2+1-2x+x^2=1$
$2x^2-2x=0$
$2x(x-1)=0$
$x=0$ or $x=1$
Step 2:
Area of the required region is shaded region shown in the figure.
Step 3:
So the required area is given by
$A=\int_0^1(y_2-y_1)dx$
$\;\;=\int_0^1(\sqrt{1-x^2}-(1-x))dx$
$\;\;\;=\int_0^1\sqrt{1-x^2}dx-\int_0^1(1-x)dx$
Step 4:
On integrating we get,
$A=\big[\large\frac{1}{2}$$x\sqrt{1-x^2}+\large\frac{1}{2}$$\sin^{-1}\big(\large\frac{x}{1}\big)$$-x+\large\frac{x^2}{2}\big]_0^1$
On applying the limits we get,
$A=[0+\large\frac{1}{2}$$\sin^{-1}(1)-1+\large\frac{1}{2}\big]$$-0$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
$A=\big(\large\frac{\pi}{4}-\frac{1}{2})$sq.units.
answered Oct 4, 2013 by sreemathi.v
 

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